Consider the sequence $n^2+3$ with $n\ge 1$.
If we write down the first terms of this sequence we get:
$4,7,12,19,28,39,52,67,84,103,124,147,172,199,228,259,292,327,364,\dots$ .
We see that the terms for $n=6$ and $n=7$ ($39$ and $52$) are both divisible by $13$.
In fact $13$ is the largest prime dividing any two successive terms of this sequence.

Let $P(k)$ be the largest prime that divides any two successive terms of the sequence $n^2+k^2$.

Find the last $18$ digits of ${\displaystyle \sum _{k=1}^{10000000}P(k)}$.